hmm I got 66. I got my radius to be 65 and from there i proceede to solve and got 2(33)=66. My method is that the radius from point O the origin perpendicular bisects the three parallel chords. Using pythagoras theorem, let the distance of perpendicular bisector from O to TU be k. Then,
k^2+36^2=(8-k)^2+25^2 and hence k=-52. After that r = sqrt(60^2 + 25^2)=65. From there sqrt(65^2-56^2)=33